derivative of y=xe^2x?

Answer 1

The derivative of uv = u'v + v'u
u = x
v = e^2x
u' = 1
v' = (the derivative of e^K = dk/dx*e^K (kind of the definition of e)) 2e^2x

= e^2x + 2xe^2x
= (2x + 1)e^2x

Answer 2

To solve this, you have to use the product rule, where, if
y = f(x) g(x)

f(x) = x and g(x) = e^(2x).

A verbal description of the product rule: "The derivative of the first times the second, plus the derivative of the second times the first."

Therefore, if

y = xe^(2x), then
y' = (1)e^(2x) + x(e^2x)[2]

NOTE: I used square brackets for the terms involved in the chain rule. In this case, we're taking the derivative of e^(2x), and then taking the derivative of the inside of the function 2x, which is 2.

Through more simplification,

y' = e^(2x) + 2xe^(2x)

And, if you really wanted to make a clean answer, you can factor that.

y' = e^(2x) [1 + 2x]

Answer 3

y=xe^2x
function 1 = x
function 2 = e^2x
derivative of function 1 = 1
derivative of function 2 = 2e^2x
y = (function 1)(derivative of function 2)+(function 2)(derivative of function 1)
y = (x)(2e^2x)+(e^2x)(1)
y=(2x+1)(e^2x)

Answer 4

(2x+1)e^(2x)